2
$\begingroup$

Conjecture: Every discontinuous utility function $U$ representing continuous preferences can be written as $U = f \circ g$ for some continuous $g$ and discontinuous strictly monotone $f$.

The goal is to prove or disprove this conjecture.

We know that continuous utility implies continuous preferences, and running it through a strict monotone gives us the same preferences but with a discontinuous utility function. The conjecture is inspired from this.

$\endgroup$
5
  • $\begingroup$ I recommend that you delete your counter-example and the edit from the body of the Q. $\endgroup$
    – Giskard
    Apr 7, 2022 at 14:12
  • $\begingroup$ Are the continuous preferences themselves monotone? $\endgroup$
    – Giskard
    Apr 7, 2022 at 14:14
  • $\begingroup$ @Giskard The preferences need not be monotone. (I have edited the post as well.) $\endgroup$ Apr 7, 2022 at 14:16
  • $\begingroup$ Any assumptions on the domain of $U$? $\endgroup$ Apr 7, 2022 at 14:36
  • $\begingroup$ @MichaelGreinecker The domain $X \subseteq \mathbb{R}^n$. But I would like to see these two cases, if possible: (1) $X \subseteq \mathbb{R}^n$ and (2) $X = \mathbb{R}^n$. $\endgroup$ Apr 7, 2022 at 14:57

2 Answers 2

3
$\begingroup$

I took the question as asking for a representation with $f:\mathbb{R}\to\mathbb{R}$. If we only require $f$ to be defined on a subset of $\mathbb{R}$, Amit's answer solves the problem.

Here is a proof for the case that the domain is connected (and second countable): Under these assumptions, there exists a continuous utility representation $V:X\to\mathbb{R}$. Since continuous functions map connected sets to connected sets, the image $V(X)$ is a, possibly unbounded, interval. I do the special case that $V(X)=[a,b)$ for some real numbers $a$ and $b$, the other cases can be dealt with by analogous methods. Now define the surjective continuous function $g:X\to [a,\infty)$ by $$g(x)=\frac{V(x)-a}{b-V(x)}.$$ Note that $g$ represents the preference ordering. Let $r^*=U(x)$ for any (and hence all) $x\in g^{-1}(a).$

Now define $f:\mathbb{R}\to\mathbb{R}$ so that for $r\geq a$ we have $f(r)=U(x)$ for any (and hence all) $x\in g^{-1}(r)$ and for $r<a$ we have $f(r)=r^*-|r-a|$. It is easy to verify that $f$ is strictly increasing and $U=f\circ g$.

$\endgroup$
1
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – 1muflon1
    Apr 10, 2022 at 12:36
4
$\begingroup$

If the preference is continuous, reflexive, transitive and complete, then there exists a continuous utility representation $g$, and since $U$ also represents the same preference, so there must be a strictly increasing function $f$ such that $U= f\circ g$.

$\endgroup$
5
  • $\begingroup$ Can you please elaborate the second part where you say that such an $f$ must exist? It may be true, but I can't convince myself that it really is the case. A proof would be appreciated! $\endgroup$ Apr 7, 2022 at 16:30
  • $\begingroup$ Take it as an exercise. Try to construct such a $f$, it is easy. $\endgroup$
    – Amit
    Apr 7, 2022 at 17:03
  • $\begingroup$ The existence of $g$ comes from Debreu's theorem. Then define $f$ such that $f(g(x)) = U(x)$. Since $g$ and $U$ denote the same preferences, it follows that $f(g(x)) > f(g(y)) \iff U(x) \geq U(y) \iff g(x) \iff g(y)$ which further implies that $f$ is monotone. Does this work? $\endgroup$ Apr 8, 2022 at 0:18
  • $\begingroup$ $f$ defined as $f(g(x)) = U(x)$ works provided you want $f:\text{Range}(g)\rightarrow\mathbb{R}$. Here $\text{Range}(g)$ will be an interval because $g$ is continuous. If you want $f:\mathbb{R}\rightarrow\mathbb{R}$, then you just need to complete the construction of $f$ outside the range of $g$. $\endgroup$
    – Amit
    Apr 8, 2022 at 2:09
  • $\begingroup$ @Pocket Cat If you want $f:\mathbb{R}\rightarrow\mathbb{R}$, see Michael's answer. $\endgroup$
    – Amit
    Apr 8, 2022 at 7:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.